Eric Bohlman <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...
> Dick Startz <[EMAIL PROTECTED]> wrote in
> news:[EMAIL PROTECTED]:
>
> > The suggestion below is the right idea, but I think is off in a
> > detail. Doing least squares we take the partial derivatives of the sum
> > of (y-b1-b2*x)^2 with respect to b1 and b2. I think the derivative
> > w.r.t x is given below by mistake.
>
> Yep, I realized that a little while after posting. Using the derivative
> with respect to b1 (2*(b1+b2*x-y)) works, though, since the subexpression
> that needs to be zero is the same.'
I think you can just show that the point (xbar,ybar) is on the line.
Since y^=b1+b2x, by summing both side from 1 to n,
Sum(y from 1 to n) = Sum[(b1+b2*(X from 1 to n)]
=nb1 + b2*[sum(x from 1 to n)]
Dividing both side by n, you get
ybar = b1 + b2*Xbar.
This shows that (xbar,ybar) is on the line :)
Let me know if you still have question.
BW
.
.
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